►zthomae’s SICP solutions
▼2 Chapter 2
2.1 Exercise 2.1
2.2 Exercise 2.2
2.3 Exercise 2.3
2.4 Exercise 2.4
2.5 Exercise 2.5
2.6 Exercise 2.6
2.7 Exercise 2.7
2.8 Exercise 2.8
2.9 Exercise 2.9
2.10 Exercise 2.10
2.11 Exercise 2.11
2.12 Exercise 2.12
2.13 Exercise 2.13
2.14 Exercise 2.14
2.15 Exercise 2.15
2.16 Exercise 2.16
2.17 Exercise 2.17
2.18 Exercise 2.18
2.19 Exercise 2.19
2.20 Exercise 2.20
2.21 Exercise 2.21
2.22 Exercise 2.22
2.23 Exercise 2.23
2.24 Exercise 2.24
2.25 Exercise 2.25
2.26 Exercise 2.26
2.27 Exercise 2.27
2.28 Exercise 2.28
2.29 Exercise 2.29
2.30 Exercise 2.30
2.31 Exercise 2.31
2.32 Exercise 2.32
2.33 Exercise 2.33
2.34 Exercise 2.34
2.35 Exercise 2.35
2.36 Exercise 2.36
2.37 Exercise 2.37
2.38 Exercise 2.38
2.39 Exercise 2.39
2.40 Exercise 2.40
2.41 Exercise 2.41
2.42 Exercise 2.42
2.43 Exercise 2.43
2.44 Exercise 2.44
2.45 Exercise 2.45
2.46 Exercise 2.46
2.47 Exercise 2.47
2.48 Exercise 2.48
2.49 Exercise 2.49
2.50 Exercise 2.50
2.51 Exercise 2.51
2.52 Exercise 2.52
2.53 Exercise 2.53
2.54 Exercise 2.54
2.55 Exercise 2.55
2.56 Exercise 2.56
2.57 Exercise 2.57
2.58 Exercise 2.58
2.59 Exercise 2.59
2.60 Exercise 2.60
2.61 Exercise 2.61
2.62 Exercise 2.62
2.63 Exercise 2.63
2.64 Exercise 2.64
2.65 Exercise 2.65
2.66 Exercise 2.66
2.67 Exercise 2.67
2.68 Exercise 2.68
2.69 Exercise 2.69
2.70 Exercise 2.70
2.71 Exercise 2.71
2.72 Exercise 2.72
2.73 Exercise 2.73
2.74 Exercise 2.74
2.75 Exercise 2.75
2.76 Exercise 2.76
2.77 Exercise 2.77
2.78 Exercise 2.78
2.79 Exercise 2.79
2.80 Exercise 2.80
2.81 Exercise 2.81
2.82 Exercise 2.82
2.83 Exercise 2.83
2.84 Exercise 2.84
2.85 Exercise 2.85
2.86 Exercise 2.86
2.87 Exercise 2.87
2.88 Exercise 2.88
2.89 Exercise 2.89
2.90 Exercise 2.90
2.91 Exercise 2.91
2.92 Exercise 2.92
2.93 Exercise 2.93
2.94 Exercise 2.94
2.95 Exercise 2.95
2.96 Exercise 2.96
2.97 Exercise 2.97
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2.8 Exercise 2.8

Alyssa reasoned that adding two intervals made a new interval where the lower bound was the sum of the arguments’ lower bounds and the upper bound was the sum of the arguments’ upper bounds. We can define a similar procedure sub-interval by supposing that subtracting two intervals creates an interval where the lower bound is the difference between the two lower bounds (and similarly for the upper bound).

(define (sub-interval x y)
  (make-interval (- (lower-bound x) (lower-bound y))
                 (- (lower-bound x) (lower-bound y))))
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